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Commit 493eb33

Browse files
feat: add solutions to lc problems: No.2810~2812 (doocs#1476)
- No.2810.Faulty Keyboard - No.2811.Check if it is Possible to Split Array - No.2812.Find the Safest Path in a Grid
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‎solution/2800-2899/2810.Faulty Keyboard/README.md‎

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@@ -158,6 +158,24 @@ function finalString(s: string): string {
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn final_string(s: String) -> String {
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let mut t = Vec::new();
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for c in s.chars() {
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if c == 'i' {
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t.reverse();
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} else {
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t.push(c);
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}
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}
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t.into_iter().collect()
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}
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}
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```
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### **...**
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```

‎solution/2800-2899/2810.Faulty Keyboard/README_EN.md‎

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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn final_string(s: String) -> String {
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let mut t = Vec::new();
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for c in s.chars() {
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if c == 'i' {
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t.reverse();
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} else {
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t.push(c);
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}
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}
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t.into_iter().collect()
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}
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}
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```
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### **...**
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```
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impl Solution {
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pub fn final_string(s: String) -> String {
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let mut t = Vec::new();
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for c in s.chars() {
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if c == 'i' {
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t.reverse();
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} else {
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t.push(c);
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}
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}
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t.into_iter().collect()
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}
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}

‎solution/2800-2899/2811.Check if it is Possible to Split Array/README.md‎

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时间复杂度 $O(n^3),ドル空间复杂度 $O(n^2)$。其中 $n$ 是数组 $nums$ 的长度。
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**方法二:脑筋急转弯**
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不论如何操作,最终总会剩下一个 `length == 2` 的子数组,又因为元素数值不存在负数,所以随着分割操作的进行,子数组的长度和总和都会逐渐变小,其它 `length > 2` 子数组之和肯定要比该子数组之和更大,进而,我们只需要考虑,是否存在一个 `length == 2` 且总和大于等于 `m` 的子数组即可。
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> 📢 注意,当 `nums.length <= 2` 时,无需进行操作。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
@@ -258,6 +266,40 @@ function canSplitArray(nums: number[], m: number): boolean {
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}
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```
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```ts
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function canSplitArray(nums: number[], m: number): boolean {
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const n = nums.length;
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if (n <= 2) {
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return true;
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}
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for (let i = 1; i < n; i++) {
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if (nums[i - 1] + nums[i] >= m) {
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return true;
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}
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}
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return false;
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn can_split_array(nums: Vec<i32>, m: i32) -> bool {
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let n = nums.len();
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if (n <= 2) {
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return true;
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}
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for i in 1..n {
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if nums[i - 1] + nums[i] >= m {
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return true;
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}
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}
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false
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}
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}
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```
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### **...**
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```

‎solution/2800-2899/2811.Check if it is Possible to Split Array/README_EN.md‎

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@@ -217,6 +217,40 @@ function canSplitArray(nums: number[], m: number): boolean {
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}
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```
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```ts
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function canSplitArray(nums: number[], m: number): boolean {
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const n = nums.length;
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if (n <= 2) {
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return true;
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}
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for (let i = 1; i < n; i++) {
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if (nums[i - 1] + nums[i] >= m) {
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return true;
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}
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}
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return false;
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn can_split_array(nums: Vec<i32>, m: i32) -> bool {
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let n = nums.len();
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if (n <= 2) {
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return true;
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}
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for i in 1..n {
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if nums[i - 1] + nums[i] >= m {
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return true;
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}
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}
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false
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}
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}
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```
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### **...**
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```
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impl Solution {
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pub fn can_split_array(nums: Vec<i32>, m: i32) -> bool {
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let n = nums.len();
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if (n <= 2) {
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return true;
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}
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for i in 1..n {
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if nums[i - 1] + nums[i] >= m {
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return true;
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}
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}
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false
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}
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}

‎solution/2800-2899/2812.Find the Safest Path in a Grid/README.md‎

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@@ -494,6 +494,132 @@ function maximumSafenessFactor(grid: number[][]): number {
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}
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```
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```ts
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function maximumSafenessFactor(grid: number[][]): number {
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const n = grid.length;
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const g = Array.from({ length: n }, () => new Array(n).fill(-1));
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const vis = Array.from({ length: n }, () => new Array(n).fill(false));
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let q: [number, number][] = [];
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for (let i = 0; i < n; i++) {
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for (let j = 0; j < n; j++) {
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if (grid[i][j] === 1) {
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q.push([i, j]);
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}
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}
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}
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let level = 0;
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while (q.length) {
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const t: [number, number][] = [];
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for (const [x, y] of q) {
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if (x < 0 || y < 0 || x === n || y === n || g[x][y] !== -1) {
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continue;
516+
}
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g[x][y] = level;
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t.push([x + 1, y]);
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t.push([x - 1, y]);
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t.push([x, y + 1]);
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t.push([x, y - 1]);
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}
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q = t;
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level++;
525+
}
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const dfs = (i: number, j: number, v: number) => {
527+
if (i < 0 || j < 0 || i === n || j === n || vis[i][j] || g[i][j] <= v) {
528+
return false;
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}
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vis[i][j] = true;
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return (
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(i === n - 1 && j === n - 1) ||
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dfs(i + 1, j, v) ||
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dfs(i, j + 1, v) ||
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dfs(i - 1, j, v) ||
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dfs(i, j - 1, v)
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);
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};
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let left = 0;
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let right = level;
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while (left < right) {
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vis.forEach(v => v.fill(false));
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const mid = (left + right) >>> 1;
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if (dfs(0, 0, mid)) {
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left = mid + 1;
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} else {
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right = mid;
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}
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}
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return right;
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}
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```
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### **Rust**
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```rust
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use std::collections::VecDeque;
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impl Solution {
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fn dfs(i: usize, j: usize, v: i32, g: &Vec<Vec<i32>>, vis: &mut Vec<Vec<bool>>) -> bool {
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if vis[i][j] || g[i][j] <= v {
562+
return false;
563+
}
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vis[i][j] = true;
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let n = g.len();
566+
i == n - 1 && j == n - 1
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|| i != 0 && Self::dfs(i - 1, j, v, g, vis)
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|| i != n - 1 && Self::dfs(i + 1, j, v, g, vis)
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|| j != 0 && Self::dfs(i, j - 1, v, g, vis)
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|| j != n - 1 && Self::dfs(i, j + 1, v, g, vis)
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}
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pub fn maximum_safeness_factor(grid: Vec<Vec<i32>>) -> i32 {
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let n = grid.len();
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let mut g = vec![vec![-1; n]; n];
576+
let mut q = VecDeque::new();
577+
for i in 0..n {
578+
for j in 0..n {
579+
if grid[i][j] == 1 {
580+
q.push_back((i, j));
581+
}
582+
}
583+
}
584+
let mut level = 0;
585+
while !q.is_empty() {
586+
let m = q.len();
587+
for _ in 0..m {
588+
let (i, j) = q.pop_front().unwrap();
589+
if g[i][j] != -1 {
590+
continue;
591+
}
592+
g[i][j] = level;
593+
if i != n - 1 {
594+
q.push_back((i + 1, j));
595+
}
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if i != 0 {
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q.push_back((i - 1, j));
598+
}
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if j != n - 1 {
600+
q.push_back((i, j + 1));
601+
}
602+
if j != 0 {
603+
q.push_back((i, j - 1));
604+
}
605+
}
606+
level += 1;
607+
}
608+
let mut left = 0;
609+
let mut right = level;
610+
while left < right {
611+
let mid = (left + right) >> 1;
612+
if Self::dfs(0, 0, mid, &g, &mut vec![vec![false; n]; n]) {
613+
left = mid + 1;
614+
} else {
615+
right = mid;
616+
}
617+
}
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right
619+
}
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}
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```
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### **...**
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```

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